Question: $\dfrac{ 6f + 6g }{ -4 } = \dfrac{ -8f - 9h }{ 8 }$ Solve for $f$.
Multiply both sides by the left denominator. $\dfrac{ 6f + 6g }{ -{4} } = \dfrac{ -8f - 9h }{ 8 }$ $-{4} \cdot \dfrac{ 6f + 6g }{ -{4} } = -{4} \cdot \dfrac{ -8f - 9h }{ 8 }$ $6f + 6g = -{4} \cdot \dfrac { -8f - 9h }{ 8 }$ Multiply both sides by the right denominator. $6f + 6g = -4 \cdot \dfrac{ -8f - 9h }{ {8} }$ ${8} \cdot \left( 6f + 6g \right) = {8} \cdot -4 \cdot \dfrac{ -8f - 9h }{ {8} }$ ${8} \cdot \left( 6f + 6g \right) = -4 \cdot \left( -8f - 9h \right)$ Distribute both sides ${8} \cdot \left( 6f + 6g \right) = -{4} \cdot \left( -8f - 9h \right)$ ${48}f + {48}g = {32}f + {36}h$ Combine $f$ terms on the left. ${48f} + 48g = {32f} + 36h$ ${16f} + 48g = 36h$ Move the $g$ term to the right. $16f + {48g} = 36h$ $16f = 36h - {48g}$ Isolate $f$ by dividing both sides by its coefficient. ${16}f = 36h - 48g$ $f = \dfrac{ 36h - 48g }{ {16} }$ All of these terms are divisible by $4$ $f = \dfrac{ {9}h - {12}g }{ {4} }$